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A and B are two points on one bank of a straight river and C, D are two other points on the other bank of river, if direction from A to B is same as that from C to D and\[AB=a\],\[\angle CAD=\alpha \],\[\angle DAB=\beta \],\[\angle CBA=\gamma \], then CD is equal to

A) \[\frac{a\sin \beta \cdot \sin \gamma }{\sin \alpha \cdot \sin (\alpha +\beta +\gamma )}\]

B) \[\frac{a\sin \alpha \cdot \sin \gamma }{\sin \beta \cdot \sin (\alpha +\beta +\gamma )}\]

C) \[\frac{a\sin \alpha \cdot \sin \beta }{\sin \gamma \cdot \sin (\alpha +\beta +\gamma )}\]

D)  None of the above

Correct Answer: B

Solution :

Clearly, \[\angle CDA=\beta ,\,\,\angle DCB=\gamma ,\,\,\angle ACB=\pi -(\alpha +\beta +\gamma )\]Using since rule in \[\Delta \,CAB,\] we get                 \[\frac{AB}{\sin (\pi -(\alpha +\beta +\gamma ))}=\frac{CA}{\sin \gamma }\] \[\Rightarrow \]               \[CA=\frac{a\cdot \sin \gamma }{\sin (\alpha +\beta +\gamma )}\] Now, using sine rule in\[\Delta \,\,CAD\], we get\[\frac{CA}{\sin \beta }=\frac{CD}{\sin \alpha }\] \[\Rightarrow \]\[CD=\frac{CA\cdot \sin \alpha }{\sin \beta }=\frac{a\sin \alpha \cdot \sin \gamma }{\sin \beta \cdot \sin (\alpha +\beta +\gamma )}\]